# Presentation of discrete upper triangular group

Let $G$ be the nilpotent Lie group consisting of matrices $$\begin{pmatrix} 1 & a_{12} & \cdots & a_{1,n}\\ 0 & 1 & \ddots & \vdots\\ \vdots & \ddots & \ddots & a_{n-1,n}\\ 0 & \cdots & 0 & 1 \end{pmatrix}$$ where $a_{ij}\in\mathbb R$.

I would like to find a presentation of the group $\Gamma=G\cap\mathrm{GL}_n\mathbb Z$.

The entries on the superdiagonal play a crucial role, in that the $n-1$ matrices with a single $1$ on the superdiagonal ($1$s on the diagonal and $0$s elsewhere) generate all of $\Gamma$.

Trying to write down a presentation, I would start by choosing $n-1$ generators, $a_{12},\dotsc,a_{n-1,n}$ (named after the matrix entries). It is also clear that we need commutation relations, like $$[\cdots[[[a_{12},a_{23}],[a_{23},a_{34}]],\ldots]\cdots]=\cdots=e.$$ (If $n=3$, this would just be $[[a_{12},a_{23}],a_{12}]=[[a_{12},a_{23}],a_{23}]=e$.) I am just not sure, though, if I explicitly need to require that, say, $$[a_{12},a_{34}]=e,$$ which seems to be directly related to the embedding I am thinking of. I would like a presentation in which any generators can be mapped to any of the matrices with a $1$ in the $(i,i+1)$ position. Is this possible?

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The Chevalley commutator relations may be what you are looking for. If the presentation was proven a long time ago, it was likely proven by Chevalley or Steinberg. I believe they describe your group for any ring R that satisfies some nice conditions (that the integers do not satisfy when n=2 or 3). Presumably one just needs to add a few relations if the low dimensional cases are actually exceptional. –  Jack Schmidt Jul 18 '12 at 17:48

There is a paper by D. Biss and S. Dasgupta (A Presentation for the Unipotent Group over Rings with Identity, J. Algebra, vol. 237, 691-707 (2001)) which gives a presentation of the group of upper triangular matrices over $R$ with diagonal entries equal to 1, where $R$ is any ring with identity.
In the case $R=\mathbb{Z}$, their generators are the same as the ones in the question above, and the relations are $$[a_{i,i+1},a_{j,j+1}] =1\ \ (i<j-1\leq n-2)$$
$$[a_{i,i+1},[a_{i,i+1},a_{i+1,i+2}]]=[a_{i+1,i+2},[a_{i,i+1},a_{i+1,i+2}]]=1$$ $$[[a_{i,i+1},a_{i+1,i+2}],[a_{i+1,i+2},a_{i+2,i+3}]]=1.$$
Are you sure about the last relation? If we choose $a_{ij}$ to be the identity with an extra $1$ in the $(i,j)$ position, we have that $[a_{12},a_{23}]=a_{13}$ and $[a_{23},a_{34}]=a_{24}$, but $[a_{13},a_{24}]=a_{14}\neq1$. –  Earthliŋ Jul 25 '12 at 2:22
But $[a_{13},a_{24}]$ is trivial (maybe you meant $[a_{13},a_{34}]=a_{14}?). The last relation boils down to $[a_{i,i+2},a_{i+1,i+3}]=1$, which is as we would expect since$\{i,i+2\}$and$\{i+1,i+3\}$are disjoint. Incidentally, I think Biss and Dasgupta show that the last relation is automatic for certain rings. – Shane O Rourke Jul 25 '12 at 10:28 I assume that when you are using$a_{ij}$to denote an element of$\Gamma$, you mean the identity matrix with an extra 1 in the$(i,j)$position. It is easy to write down a presentation of$\Gamma$on the$m := n(n-1)/2$generators$a_{ij}$with$i<j$. Just use the$m(m-1)/2$commutator relations$[a_{ij},a_{jk}] = a_{ik}$, and$[a_{ij},a_{kl}]=1$when$j \ne k$,$i \le k$, and$j<l$when$i=k$. Of course many of these relations are redundant. To get a presentation on the generators$a_{i,i+1}$, you can then use the equations$[a_{ij},a_{jk}] = a_{ik}$to eliminate all of the other generators. Doing that with$n=3$gives the presentation you have written down. Although many relations in this presentation are redundant, those like$[a_{12},a_{34}]=1$between the commuting generators$a_{i,i+1}$are definitely not redundant, because if you left one of those out, then you could make the group bigger by putting that commutator equal to a new central generator. - Thank you. I had hoped that one could embed$\Gamma$as abstract group into$G$by sending the generators$a_{i,i+1}$to any element in$G$as long as the entries on the superdiagonal were linearly independent. Do you see any way of understanding of how one is allowed to embed$\Gamma$into$G$? – Earthliŋ Jul 19 '12 at 1:55 You certainly cannot send the$a_{i,i+1}$to an arbitrary sequence of elements of$G$with linearly independent entries on the superdiagonal. For example, for$n=4$, you cannot have$(a_{12},a_{23},a_{34}) \to (a_{23},a_{12},a_{34})$because$[a_{12},a_{34}]= 1$but$[a_{23},a_{34}]\ne 1$. I am pretty sure that the only embeddings possible are those that either map each$a_{i,i+1}$to a "multiple" of$a_{i,i+1}$(i.e. the image can have any nonzero entry in the$(i,i+1)$-position) or those that do the same but reverse the order of the$a_{i,i+1}$. – Derek Holt Jul 19 '12 at 9:13 Interesting, I hope you are wrong. For$n=3$, the images of$a_{12}$and$a_{23}$must not commute, which is the case precisely when they map to any two elements in$G$, as long as the values on the superdiagonal are linearly independent. For general$n$the relations are not symmetric on the generators, which complicates things. – Earthliŋ Jul 19 '12 at 14:58 Maybe 'neighbouring' generators should be allowed to vary inside$\mathrm{GL}_2\mathbb R$...? – Earthliŋ Jul 19 '12 at 15:06 Yes, what I said was wrong for$n=3$, but I still believe it is true for$n>3$. Since I don't have the energy to try and write down a proof at present, let me challenge you to try and find a counterexample to my claim, say for$n=4\$. –  Derek Holt Jul 19 '12 at 15:40